Measurement of Power MCQ Quiz in मल्याळम - Objective Question with Answer for Measurement of Power - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 18, 2025
Latest Measurement of Power MCQ Objective Questions
Top Measurement of Power MCQ Objective Questions
Measurement of Power Question 1:
In the measurement of 3-phase power using two wattmeter method, the value of phase angle when the two wattmeter reads equal reading is:
Answer (Detailed Solution Below)
Measurement of Power Question 1 Detailed Solution
Power measurement using two wattmeter method
According to Blondel's Theorem, for 'n' phase power measurement, the no. of wattmeters required is 'n-1'.
Therefore, for the 3-phase circuit, 3-1 = 2 wattmeters are required.
The total 3ϕ power is the sum of the power measured by individual wattmeters.
\(W=W_1+W_2\)
The reading of the wattmeters is given by:
\(W_1=V_LI_Lcos(30+ϕ)\)
\(W_2=V_LI_Lcos(30-ϕ)\)
The total power is given by:
\(W_1+W_2=\sqrt{3}V_LI_Lcosϕ\)
Calculation
The value of phase angle (ϕ) is given by:
\(ϕ=tan^{-1}[{\sqrt{3}(W_1-W_2)\over W_1+W_2}]\)
If W1 = W2
\(ϕ=tan^{-1}(0)\)
ϕ = 0°
Important Points
p.f. angle (ϕ) |
p.f.(cos ϕ) |
W1 [VLIL cos (30 + ϕ)] |
W2 [VLIL cos (30 - ϕ)] |
W = W1 + W2 [W = √3VLIL cos ϕ] |
Observations |
0° |
1 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
√3 VLIL |
W1 = W2 |
30° |
0.866 |
\(\frac{{{V_L}I}}{2}\) |
VLIL |
1.5 VL IL |
W2 = 2W1 |
60° |
0.5 |
0 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
W1 = 0 |
90° |
0 |
\(\frac{{ - {V_L}{I_L}}}{2}\) |
\(\frac{{{V_L}{I_L}}}{2}\) |
0 |
W1 = -ve W2 = +ve |
Measurement of Power Question 2:
In case of the two-wattmeter method for measuring the power of a balanced 3ϕ load, the whole power is measured by one of the two wattmeters when:
Answer (Detailed Solution Below)
Measurement of Power Question 2 Detailed Solution
In two wattmeter method, the wattmeter readings are given by
W1 = VLIL cos (30 – ϕ)
W2 = VLIL cos (30 + ϕ)
Total power = W1 + W2
Reactive power = √3 (W1 – W2)
Power factor = cos ϕ
Where \(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{\left( {{W_1} + {W_2}} \right)}}} \right)\)
Important Point:
p.f. angle (ϕ) |
p.f.(cos ϕ) |
W1 [VLIL cos (30 + ϕ)] |
W2 [VLILcos (30 - ϕ)] |
W = W1 + W2 [W = √3VLIL cos ϕ] |
Observations |
0° |
1 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
√3 VLIL |
W1 = W2 |
30° |
0.866 |
\(\frac{{{V_L}I}}{2}\) |
VLIL |
1.5 VL IL |
W2 = 2W1 |
60° |
0.5 |
0 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
W1 = 0 |
90° |
0 |
\(\frac{{ - {V_L}{I_L}}}{2}\) |
\(\frac{{{V_L}{I_L}}}{2}\) |
0 |
W1 = -ve W2 = +ve |
Measurement of Power Question 3:
The wattmeter measures the angle between the current phasor detected by the _________ and the voltage phasor detected by the ____________.
Answer (Detailed Solution Below)
Measurement of Power Question 3 Detailed Solution
In wattmeter, the arrangement of the current coil is in series with the load and the arrangement of the potential coil is across the load.
Hence the wattmeter measures the angle between the current phasor detected by the current coil and the voltage phasor detected by the voltage coil.
The wattmeter measures the average power and it is given by
\(P = {V_{ph}}{I_{ph}}\cos \left( \phi \right)\)
ϕ is the angle between Vph and Iph
Measurement of Power Question 4:
In a three-phase measurement by two-wattmeter method, the readings of the wattmeters are equal in magnitude but opposite in the sign then the nature of the load will be
Answer (Detailed Solution Below)
Measurement of Power Question 4 Detailed Solution
Two wattmeter method:
Two wattmeter method is used to measure the power in three-phase circuits for both balanced and unbalanced load.
The wattmeter readings are
\({W_1} = {V_L}{I_L}\cos \left( {30 - ϕ } \right)\)
\( {W_2} = {V_L}{I_L}\cos \left( {30 + ϕ } \right)\)
\({W_1} + {W_2} = {V_L}{I_L}\left[ {\cos \left( {30 - ϕ } \right) + \cos \left( {30 + ϕ } \right)} \right]\)
\(= \sqrt 3 {V_L}{I_L}\cos ϕ\)
∴Total three-phase power is
\( W= {\rm{\;}}{{\rm{W}}_1} + {{\rm{W}}_2} = \sqrt 3 {V_L}{I_L}\cos ϕ \)
Total three-phase power is the sum of two wattmeters.
\({W_1} = {V_L}{I_L}\cos \left( {30 - ϕ } \right)\)
\({W_2} = {V_L}{I_L}\cos \left( {30 + ϕ } \right)\)
\({W_1} - {W_2} = \sqrt 3 {V_{ph}}{I_{ph}}\sin ϕ\)
\(\sqrt 3 \left( {{W_1} - {W_2}} \right) = 3{V_{ph}}{I_{ph}}\sin ϕ \)
Reactive power \(= \surd 3\;\left( {{W_1}-{W_2}} \right)\)
Reactive power is equal to √3 times the difference between the readings of the two wattmeters.
\({W_1} + {W_2} = 3{V_{ph}}{I_{ph}}\cos ϕ\)
\(\sqrt 3 \left( {{W_1} - {W_2}} \right) = 3{V_{ph}}{I_{ph}}\cos ϕ \)
\(\Rightarrow ϕ = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
Power factor \(= cos\;ϕ\)
\(cosϕ = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]\)
Nature of the load based on wattmeter readings:
The nature of the load w.r.t the readings shown in the below table.
Wattmeter readings |
Power factor value |
Nature of the load |
|
W1 |
W2 |
||
W1 = P |
W2 = -P |
Cos ϕ = 0 (ZPF) |
Pure inductive load |
W1 = P |
W2 = P |
Cos ϕ = 1 (UPF) |
Pure resistive load |
W1 = P |
W2 = 0 |
cos ϕ = 0.5 lag |
R-L load |
P = Active power value in Watts
Points to remember:
The value 0 < cos ϕ < 0.5 or 60° < ϕ < 90° means that one of the wattmeters reads negative value.
Measurement of Power Question 5:
In a low power factor wattmeter the pressure coil is connected
Answer (Detailed Solution Below)
Measurement of Power Question 5 Detailed Solution
- In the circuit shown in figure (a), the pressure coil is connected on the supply side of the current coil; The same current passes through the load and current coil of the wattmeter
- But the voltage across the pressure coil is more than that across the load by an amount equal to the voltage drop in the current coil
- This connection is used under light loads (or) where the load impedance is high
- This connection is used for high power factor circuits
- In the circuit shown in figure (b), the pressure coil is connected on the load side of the current coil
- The voltage across the pressure coil is the same as that across the load, but the current in the current coil is more than in the load by an amount equal to the current in the pressure coil
- This connection is used under heavy loads (or) where the load impedance is low
- This connection is used for low power factor circuits
Measurement of Power Question 6:
In a 3-phase system, two-wattmeter method is used to measure the power. If one of the wattmeters shows a negative reading and the other shows a positive reading, and the magnitude of the readings are not the same, then what will be the power factor (p.f.) of the load?
Answer (Detailed Solution Below)
Measurement of Power Question 6 Detailed Solution
Concept:
In a two-wattmeter method,
The reading of first wattmeter (W1) = VL IL cos (30 – ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 + ϕ)
Total power in the circuit (P) = W1 + W2
Total reactive power in the circuit \(Q = \sqrt 3 \left( {{W_1} - {W_2}} \right)\)
Power factor = cos ϕ
\(ϕ = {\rm{ta}}{{\rm{n}}^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
p.f. angle (ϕ) |
p.f.(cos ϕ) |
W1 [VLIL cos (30 - ϕ)] |
W2 [VLIL cos (30 + ϕ)] |
W = W1 + W2 [W = √3VLIL cos ϕ] |
Observations |
0° |
1 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
√3 VLIL |
W1 = W2 |
30° |
0.866 |
\(\frac{{{V_L}I_L}}{2}\) |
VLIL |
1.5 VL IL |
W1= 2W2 |
60° |
0.5 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
0 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
W2 = 0 |
90° |
0 |
\(\frac{{ - {V_L}{I_L}}}{2}\) |
\(\frac{{{V_L}{I_L}}}{2}\) |
0 |
W2 = -ve W1 = +ve |
Explanation:
W1 = -ve, W2 = +ve
|W1| ≠ |W2|, then the powere factor range is given as
The range of ϕ is, 60 < ϕ < 90
Now the range of power factor is, 0.5 > cos ϕ > 0.
Measurement of Power Question 7:
The current coil of a 200 V, 5 A, electrodynamometer type LPF wattmeter carries a current of \(\sqrt 2 {\rm{cos}}\left( {100\pi t} \right)\) A. The voltage across the pressure coil is \(\sqrt 2 {\rm{sin}}\left( {100\pi t} \right)\) V. The meter will indicate
Answer (Detailed Solution Below)
Measurement of Power Question 7 Detailed Solution
Concept:
The power triangle is shown below.
P = Active power (or) Real power in W = Vrms Irms cos ϕ
Q = Reactive power in VAR = Vrms Irms sin ϕ
S = Apparent power in VA = Vrms Irms
S = P + j Q
\(S = \sqrt {{P^2} + {Q^2}} \)
ϕ is the phase difference between the voltage and current
Power factor \(\cos \phi = \frac{P}{S}\)
Calculation:
Given that,
\(\begin{array}{l} V = \sqrt 2 sin\;\left( {100\pi t} \right)\\ I = \sqrt 2 cos\;\left( {100\pi t} \right) \end{array}\)
The phase difference between V and I is = 90°
Power = VI cos ϕ
P = 0 WMeasurement of Power Question 8:
In the two-wattmeter method of three phase power measurement of a balanced load, if the reading of one metre is – 200 W, then then power factor of the load is ______.
Answer (Detailed Solution Below)
Measurement of Power Question 8 Detailed Solution
In two wattmeter method, the power factor is given by
\(cos\phi = \cos \left[ {{{\tan }^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)} \right]\)
Here, \(\phi = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}} \right)\)
Given that, one reading is negative.
Hence the term (W1 + W2) is always lesser than (W1 - W2).
Now the minimum value of the term \(\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}\) is √3 and it occurs at one of the reading is zero. Here no reading is zero. Hence the value should be greater than √3 and hence ϕ will be greater than 60°.
Now the maximum value of the term \(\frac{{\sqrt 3 \left( {{W_1} - {W_2}} \right)}}{{{W_1} + {W_2}}}\) is infinite and it occurs at (W1 = -W2). In this case ϕ will be 90°.
Hence ϕ varies between 60 to 90 degrees.
Now the range of ϕ is, \(60^\circ < \phi < 90^\circ \)
Now the range of power factor is, \(0.5 > \cos \phi > 0\).
The power factor of the system is greater than or equal to 0 but less than 0.5.
Measurement of Power Question 9:
The false statement regarding Two Wattmeter method is:
Answer (Detailed Solution Below)
Measurement of Power Question 9 Detailed Solution
Two-wattmeter method:
- If one of the wattmeters shows -ve value then this indicates that the load is reactive.
- Element i.e. an inductor or a capacitor that consumes the reactive power.
- When p.f. angle = 60°, only a single wattmeter measures the entire 3-phase power.
- When p.f. angle = 60°, only one wattmeter shows zero deflection.
- When p.f. angle > 60°, one of the wattmeters shown -ve deflection.
- When p.f. angle < 60°, one of the wattmeters shown +ve deflection.
- When one of the wattmeters shows -ve deflection, the range of p.f. is from 0 to 0.5.
- When two wattmeter show +ve deflection, then a range of the p.f. is from 0 to 1.
Important Points
p.f. angle (ϕ) |
p.f.(cos ϕ) |
W1 [VLIL cos (30 + ϕ)] |
W2 [VLIL cos (30 - ϕ)] |
W = W1 + W2 [W = √3VLIL cos ϕ] |
Observations |
0° |
1 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
√3 VLIL |
W1 = W2 |
30° |
0.866 |
\(\frac{{{V_L}I}}{2}\) |
VLIL |
1.5 VL IL |
W2 = 2W1 |
60° |
0.5 |
0 |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
\(\frac{{\sqrt 3 }}{2}{V_L}{I_L}\) |
W1 = 0 |
90° |
0 |
\(\frac{{ - {V_L}{I_L}}}{2}\) |
\(\frac{{{V_L}{I_L}}}{2}\) |
0 |
W1 = -ve W2 = +ve |
Measurement of Power Question 10:
Which condition is created by two wattmeters W1 and W2 for power factor is zero?
Answer (Detailed Solution Below)
Measurement of Power Question 10 Detailed Solution
Measurement of power
According to Blondel's Theorem, for the measurement of power in n-phase, n-wire without neutral, (n - 1) wattmeters are required.
For measurement of power in 3-phase, (3 - 1 = 2) wattmeters are required.
The readings of both the wattmeters are given by:
\(W_1=V_LI_Lcos(30+ϕ)\)
\(W_2=V_LI_Lcos(30-ϕ)\)
The power factor is given by:
Power factor = cosϕ
At zero power, cosϕ = 0
At, ϕ = 90°, the readings of both the wattmeters are:
\(W_1=V_LI_Lcos(30+90)={-0.5V_LI_L}\)
\(W_2=V_LI_Lcos(30-90)={+0.5V_LI_L}\)
Thus, W1 = -ve, W2 = +Ve
Thus at zero power factor, the reading of both wattmeters are equal and opposite.